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A particle moves along a straight line such that its displacement at any time t is given by s = t 3 - 6t 2 + 3t + 4 metres The velocity when the acceleration is zero is (a) 3 ms -1 (b) -12 ms -1 (c) 42 ms -2 (d) -9 ms -1



A particle moves along a straight line such that its displacement at any time t is given by


s = t3 - 6t2 + 3t + 4 metres


The velocity when the acceleration is zero is


(a)    3 ms-1       


(b)    -12 ms-1


        (c)    42 ms-2     


(d)    -9 ms-1


Grade:10

7 Answers

Amit Saxena
35 Points
9 years ago

(d)
Velocity, v=  ds/dt=3t^2  -12 t+3
Acceleration a = ds/dt = 6t - 12; For a = 0 we have, 0 = 6 t - 12 or t= 2s. Hence, at t = 2 s the velocity will be
v = 3 × 22 – 12 × 2 + 3 = -9 ms-1

Shubh Prakash
13 Points
5 years ago
(d) -9m/s we know that differentiation of displacement gives velocity i.e. 3t^2-12t+3.now we know that differentiation of velocity gives acceleration i.e.6t-12but here acceleration is zero;therefore,6t-12=0;t=2secnow put the value of `t` in velocity we get -9m/sThank You
Sandeep kaur
15 Points
5 years ago
Your method is correct but your calculation is not correct .the answer is -15m/s .so correct your answer both of you and thank you .
Vineet Vilas koli
13 Points
4 years ago
ds/dt=v=3t^2-12t+3
d^2s/dt^2=a=6t-12
since a=0
therefore, 6t-12=0
                   6t=12
                      t=2
        put t=2 in v
         v= 3t^2-12t
         v=-12m/s
           =3(2^2)-12×2
         V= -12m/s
SALAJAR
17 Points
4 years ago
Vineet Vilas kol …. BRO YOU HAVE BEEN PUT YOUR LEG IN DAL BRO YOU LEFT THE NUMBER 3 ON YOUR CALCULATION... SO CHECK IT ONCE.... AND THE CORRECT THE ANSWER TO THE OPTION D (-9)
 
Ajeet Tiwari
askIITians Faculty 86 Points
3 years ago
hello student

since ,Velocity, v= ds/dt=3t^2 -12 t+3
Acceleration a = ds/dt = 6t - 12; For a = 0 we have, 0 = 6 t - 12 or t= 2s. Hence, at t = 2 s the velocity will be
v = 3 × 2^2 – 12 × 2 + 3 = -9 ms-1

Hope it helps
Thankyou
Yash Chourasiya
askIITians Faculty 256 Points
3 years ago
Dear Student

Given
s= t3 − 6t2 + 3t + 4
v = ds/dt​
So v = 3t2 −12t + 3.............(1)
again differentiate, a = dv/dt​
6t − 12=0
t = 2s
Putting value of t in eq. (1)
v = 3(2)2 −12×2 +3
v = −9ms−1

I hope this answer will help you.
Thanks & Regards
Yash Chourasiya

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